POJ1833 & POJ3187 & POJ3785 next_permutation应用

Posted 2020-09-23 gccbuaa

要是没有next_permutation这个函数，这些题认为还不算特别水，只是也不一定，那样可能就会有对应的模板了。

反正正是由于next_permutation这个函数。这些题包含之前的POJ1226，都变得简单起来。

排列

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 17486		Accepted: 6970

Description

题目描写叙述： 
大家知道，给出正整数n，则1到n这n个数能够构成n。种排列，把这些排列依照从小到大的顺序（字典顺序）列出，如n=3时，列出1 2 3，1 3 2，2 1 3，2 3 1，3 1 2，3 2 1六个排列。 

任务描写叙述： 
给出某个排列。求出这个排列的下k个排列，假设遇到最后一个排列，则下1排列为第1个排列，即排列1 2 3…n。

 
比方：n = 3，k=2 给出排列2 3 1，则它的下1个排列为3 1 2，下2个排列为3 2 1，因此答案为3 2 1。 

Input

第一行是一个正整数m。表示測试数据的个数，以下是m组測试数据。每组測试数据第一行是2个正整数n( 1 <= n < 1024 )和k(1<=k<=64)，第二行有n个正整数，是1，2 … n的一个排列。

Output

对于每组输入数据，输出一行。n个数，中间用空格隔开，表示输入排列的下k个排列。

Sample Input

3
3 1
2 3 1
3 1
3 2 1
10 2	
1 2 3 4 5 6 7 8 9 10

Sample Output

3 1 2
1 2 3
1 2 3 4 5 6 7 9 8 10

直接用next_permutation这个函数就可以。

代码：

#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
#include <string>
#include <cstring>
using namespace std;

int num[1030];

int main()
{
	int Test,N,Q,i;
	cin>>Test;
	while(Test--)
	{
		scanf_s("%d%d",&N,&Q);
		for(i=0;i<N;i++)
			scanf_s("%d",&num[i]);
		for(i=1;i<=Q;i++)
			 next_permutation(num,num+N);
		for(i=0;i<N;i++)
			printf("%d ",num[i]);
		printf("\n");
	}
	return 0;
}

Backward Digit Sums

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5072		Accepted: 2923

Description

FJ and his cows enjoy playing a mental game. They write down the numbers from 1 to N (1 <= N <= 10) in a certain order and then sum adjacent numbers to produce a new list with one fewer number. They repeat this until only a single number is left. For example, one instance of the game (when N=4) might go like this: 

    3   1   2   4

      4   3   6

        7   9

         16

Behind FJ‘s back, the cows have started playing a more difficult game, in which they try to determine the starting sequence from only the final total and the number N. Unfortunately, the game is a bit above FJ‘s mental arithmetic capabilities. 

Write a program to help FJ play the game and keep up with the cows.

Input

Line 1: Two space-separated integers: N and the final sum.

Output

Line 1: An ordering of the integers 1..N that leads to the given sum. If there are multiple solutions, choose the one that is lexicographically least, i.e., that puts smaller numbers first.

Sample Input

4 16

Sample Output

3 1 2 4

Hint

Explanation of the sample: 

There are other possible sequences, such as 3 2 1 4, but 3 1 2 4 is the lexicographically smallest.

题意是要输出N个数，这N个数是从1到N这些数的一个顺序。这种顺序依照杨辉三角的模式相加起来等于sum，输出相等时的第一个字典顺序。

一看到N是大于1小于10的我就想暴力了。。

。

代码：

#include <iostream>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;

int main()
{
	int i,n,sum,a[12];
	cin>>n>>sum;

	for(i=1;i<=10;i++)
		a[i]=i;
	if(n==1)
	{
		cout<<1<<endl;
	}
	else if(n==2)
	{
		cout<<1<<" "<<2<<endl;
	}
	else if(n==3)
	{
		while(1*a[1]+2*a[2]+1*a[3]!=sum)
		{
			next_permutation(a+1,a+3+1);
		}
		cout<<a[1]<<" "<<a[2]<<" "<<a[3]<<endl;
	}
	else if(n==4)
	{
		while(1*a[1]+3*a[2]+3*a[3]+1*a[4]!=sum)
		{
			next_permutation(a+1,a+4+1);
		}
		cout<<a[1]<<" "<<a[2]<<" "<<a[3]<<" "<<a[4]<<endl;
	}
	else if(n==5)
	{
		while(1*a[1]+4*a[2]+6*a[3]+4*a[4]+1*a[5]!=sum)
		{
			next_permutation(a+1,a+5+1);
		}
		cout<<a[1]<<" "<<a[2]<<" "<<a[3]<<" "<<a[4]<<" "<<a[5]<<endl;
	}
	else if(n==6)
	{
		while(1*a[1]+5*a[2]+10*a[3]+10*a[4]+5*a[5]+1*a[6]!=sum)
		{
			next_permutation(a+1,a+n+1);
		}
		cout<<a[1]<<" "<<a[2]<<" "<<a[3]<<" "<<a[4]<<" "<<a[5]<<" "<<a[6]<<endl;
	}
	else if(n==7)
	{
		while(1*a[1]+6*a[2]+15*a[3]+20*a[4]+15*a[5]+6*a[6]+1*a[7]!=sum)
		{
			next_permutation(a+1,a+n+1);
		}
		cout<<a[1]<<" "<<a[2]<<" "<<a[3]<<" "<<a[4]<<" "<<a[5]<<" "<<a[6]<<" "<<a[7]<<endl;
	}
	else if(n==8)
	{
		while(1*a[1]+7*a[2]+21*a[3]+35*a[4]+35*a[5]+21*a[6]+7*a[7]+1*a[8]!=sum)
		{
			next_permutation(a+1,a+n+1);
		}
		cout<<a[1]<<" "<<a[2]<<" "<<a[3]<<" "<<a[4]<<" "<<a[5]<<" "<<a[6]<<" "<<a[7]<<" "<<a[8]<<endl;
	}
	else if(n==9)
	{
		while(1*a[1]+8*a[2]+28*a[3]+56*a[4]+70*a[5]+56*a[6]+28*a[7]+8*a[8]+1*a[9]!=sum)
		{
			next_permutation(a+1,a+n+1);
		}
		cout<<a[1]<<" "<<a[2]<<" "<<a[3]<<" "<<a[4]<<" "<<a[5]<<" "<<a[6]<<" "<<a[7]<<" "<<a[8]<<" "<<a[9]<<endl;
	}
	else if(n==10)
	{
		while(1*a[1]+9*a[2]+36*a[3]+84*a[4]+126*a[5]+126*a[6]+84*a[7]+36*a[8]+9*a[9]+1*a[10]!=sum)
		{
			next_permutation(a+1,a+n+1);
		}
		cout<<a[1]<<" "<<a[2]<<" "<<a[3]<<" "<<a[4]<<" "<<a[5]<<" "<<a[6]<<" "<<a[7]<<" "<<a[8]<<" "<<a[9]<<" "<<a[10]<<endl;
	}

	return 0;
}

The Next Permutation

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 979		Accepted: 717

Description

For this problem, you will write a program that takes a (possibly long) string of decimal digits, and outputs the permutation of those decimal digits that has the next larger value (as a decimal number) than the input number. For example: 

123 -> 132 
279134399742 -> 279134423799 

It is possible that no permutation of the input digits has a larger value. For example, 987.

Input

The first line of input contains a single integer P, (1 ≤ P ≤ 1000), which is the number of data sets that follow. Each data set is a single line that contains the data set number, followed by a space, followed by up to 80 decimal digits which is the input value.

Output

For each data set there is one line of output. If there is no larger permutation of the input digits, the output should be the data set number followed by a single space, followed by the string BIGGEST. If there is a solution, the output should be the data set number, a single space and the next larger permutation of the input digits.

Sample Input

3 
1 123 
2 279134399742 
3 987

Sample Output

1 132
2 279134423799
3 BIGGEST

还是直接使用next_permutation。这个函数是有返回值的，返回值是0时表示已经没有下一个字典顺序了，它要变成第一个字典顺序。返回值是1时表示还有字典顺序的下一个顺序，所以利用函数的这个性质就OK了。

代码：

#include <iostream>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;

int main()
{
	int Test,num;
	char s[100];

	cin>>Test;
	while(Test--)
	{
		cin>>num>>s;
		cout<<num<<" ";

		int n=next_permutation(s,s+strlen(s));
		
		if(n==0)
			cout<<"BIGGEST"<<endl;
		else
			cout<<s<<endl;
	}
	return 0;
}